Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x - 8} = \dfrac{x + 1}{x - 8}$
Multiply both sides by $x - 8$ $ \dfrac{x^2 - 1}{x - 8} (x - 8) = \dfrac{x + 1}{x - 8} (x - 8)$ $ x^2 - 1 = x + 1$ Subtract $x + 1$ from both sides: $ x^2 - 1 - (x + 1) = x + 1 - (x + 1)$ $ x^2 - 1 - x - 1 = 0$ $ x^2 - 2 - x = 0$ Factor the expression: $ (x + 1)(x - 2) = 0$ Therefore $x = -1$ or $x = 2$ The original expression is defined at $x = -1$ and $x = 2$, so there are no extraneous solutions.